Question

The half-life of a radioactive element is $5$ years. The fraction of the radioactive substance that remains after $20$ years is:

• A. $1/8$
• B. $1/4$
• C. $1/2$
• D. $1/16$

Answer 1

The correct option is D $1/16$

Let $A_o$ be initial concentration

$t_{1/2}=5years$

So, after $5$ years, concentration remaining is $\frac{A_o}{2}$

For first order reaction, we know

$k=rateconstant=\frac{0.693}{t_{1/2}}$

$\Rightarrow k=\frac{0.693}{5}$

Now, let $A_o$ is initial concentration and $A_{20}$ be concentration after $20$ years.

For first order reaction, we know

$k=\frac{2.303}{t}lo{g}_{10}\left(\frac{A_o}{A_t}\right)$

$\Rightarrow \frac{0.693}{5}=\frac{2.303}{20}lo{g}_{10}\left(\frac{A_o}{A_{20}}\right)$

$\Rightarrow lo{g}_{10}\left(\frac{A_0}{A_{20}}\right)=\frac{0.693\times 20}{5\times 2.303}$

$\Rightarrow \frac{A_0}{A_{20}}=(10{)}^{\frac{0.693\times 20}{5\times 2.303}}$

$\Rightarrow \frac{A_0}{A_{20}}=16$

$\Rightarrow A_{20}=\frac{A_0}{16}$

$\therefore$ Fraction remaining $=\frac{1}{16}$